求x*x y*y z*z=1上平行于平面x-y 2z=0的切线方程
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(x+y+z)^2=25x^2+y^2+z^2+2*(x+y+z)=25z^2=23-(x^2+Y^2)0
(x+y+z)²=1,x²+2xy+y²+2(x+y)z+z²=1,x²+y²+z²+2(x+y)z+2xy=1xy+yz+xz=
x-3=y-2x-y=1y-2=z-1y-z=1x-3=z-1z-x=-2x^2+y^2+z^2-xy-yz-xz=x(x-y)+y(y-z)+z(z-x)=x+y-2zx-3=z-1y-2=z-12
dz=(∂z/∂x)dx+(∂z/∂y)dyxy+yz+xz-1=0设g(x,y,z)=xy+yz+xz-1 ∂g/∂x=y+
首先,显然x,y,z均不为0.然后分开看xy:yz=3:2,两边除以y,得x:z=3:2yz:zx=2:1,除以z,得y:x=2:1,两边同时乘以3,得x:y=3:6所以:x:y:z=3:6:2,不能
由xy/(x+y)=1,yz/(y+z)=2,zx/(z+x)=3,得:(x+y)/xy=1,(y+z)/yz=1/2,(z+x)/zx=1/3,(取倒数)所以1/x+1/y=1,(1)1/y+1/z
xyz=1所以z=1/xyxz=1/yyz=1/xx/(xy+x+1)+y/(yz+y+1)+z/(xz+z+1)=x/(xy+x+1)+y/(1/x+y+1)+(1/xy)/(1/y+1/xy+1)
本题考查最值不等式:a+b≥2√ab当且仅当a=b时,取等号x√yz+y√zx+z√xy≤x(y+z)/2+y(z+x)/2+z(x+y)/2当且仅当y=z,z=x,x=y,即:x=y=z时,取等号,
y=-12;一共是三个方程,因为xy/(x+y)=3推出(x+y)/(xy)=1/3-------方程1;同理:(y+z)/(yz)=1/2-------方程2;(x+z)/(xz)=1-------
(x+y+z)²=1²x²+y²+z²+2xy+2yz+2xz=1x²+y²+z²+2(xy+yz+xz)=1x&sup
这是道竞赛题我在电脑前没有笔,所以无法给出正确结果,但可以给你思路设f(t)=(t-x)(t-y)(t-z)则f(t)=t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz代入x+y+z=1,
1/Y+1/X=1(1)1/Z+1/Y=2(2)1/X+1/Z=3(3)(1)+(2)+(3):1/X+1/Y+1/Z=3(4)(4)-(1):1/Z=2Z=1/2(4)-(2):1/X=1X=1题目
x^2+y^2+z^2+2(xy+yz+zx)=(x+y+z)^2=1由柯西不等式有x^2+y^2+z^2>=(x+y+z)^2/3=1/3所以xy+yz+zx=(1-x^2-y^2-z^2)/2
图片中的题可以用琴森不等式构造函数f(x)=e^x/(3e^x+1)^0.5可以验证f``(x)>0对所有x成立因此f(x)是下凸函数有f(x)+f(y)+f(z)>=3f(x+y+z/3)令x=ln
同学,xyz=1吧?这样的话,原式=x/(xy+x+xyz)+y/(yz+y+xyz)+z/(xz+z+xyz)=1/(y+1+yz)+1/(z+1+xz)+1/(x+1+xy)=xyz/(y+xyz
|x-3|+|y+z|+|2z+1|=0则|x-3|=0x=3|y+z|=0y=-z=1/2|2z+1|=0z=-1/2xy-yz=3x1/2-1/2x(-1/2)=7/4
(x+y+z)²=x²+y²+z²+2xy+2yz+2xz所以可得:xy+yz+xz=[(x+y+z)²-(x²+y²+z
xy:yz:zx=3:2:1xy:yz=3:2则x:z=3:2同理y:z=3:1=6:2故(x+y):z=(3+6):2=9:2
①x:y:z因为xy:yz:zx=3:2:1所以xy:yz=3:2所以x:z=3:2同理yz:zx=2:1所以y:x=2:1=6:3所以x:y:z=3:6:2②x/yz:y/zx=x^2:y^2=(x
解x^2+y^2+z^2=1x^2+(y/根2)^2+(y/根2)^2+z^2=12xy/根2+2yz/根2